inequality

• Jan 26th 2013, 12:09 PM
mulfer
inequality
• Jan 26th 2013, 10:35 PM
earthboy
Re: inequality
Quote:

Originally Posted by mulfer

We have to prove:
\$\displaystyle a^2+b^2+c^2 \geq a+b+c\$
It is equivalent to proving:
\$\displaystyle a^2+b^2+c^2-a-b-c \geq 0\$
\$\displaystyle \implies a(a-1)+b(b-1)c(c-1) \geq 0 \$ So we are done if we are able to prove this.
now,
if \$\displaystyle x\$ is a positive real number,
if\$\displaystyle x>1\$ then \$\displaystyle x(x-1)>(x-1)\$ (obvious)
if \$\displaystyle x=1 \$then \$\displaystyle x(x-1)=(x-1) \$(obvious)
if \$\displaystyle 0<x<1\$ then \$\displaystyle (x-1)<0 \$ so, \$\displaystyle x<1\$ \$\displaystyle \implies x(x-1)>x-1\$ (multiplying both sides by a negative number i.e \$\displaystyle (x-1)\$ changes the sign)
so for all case, \$\displaystyle x(x-1)\geq x-1\$
similarly,
\$\displaystyle a(a-1) \geq a-1\$
\$\displaystyle b(b-1) \geq b-1\$
\$\displaystyle c(c-1) \geq c-1\$
summing these we get:
\$\displaystyle a(a-1)+b(b-1)c(c-1) \geq (a-1)+(b-1)+(c-1) = (a+b+c)-3\$
so we will be done if we prove \$\displaystyle (a+b+c)-3 \geq 0\$ or \$\displaystyle a+b+c \geq 3\$
by the AM-GM inequality,
\$\displaystyle (a+b+c)\geq 3(abc)^{1/3}=3\$ # as \$\displaystyle abc=1\$
proved (Rock)
• Jan 27th 2013, 04:14 AM
mulfer
Re: inequality